Let \( x=at^2,y=2at \) . If \( \displaystyle\frac{d^2y}{dx^2}+\displaystyle\frac{dy}{dx}=0 \) at \( t=3 \) , then \( a \) is equal to:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{1}{18} \)
By parametric differentiation:
\( \displaystyle \frac{dy}{dx}=\frac{dy/dt}{dx/dt} \)
\( \displaystyle =\frac{2a}{2at}=\frac{1}{t} \)
For the second derivative, \( \frac{dy}{dx} \) is differentiated with respect to t and divided by dx/dt again:
\( \displaystyle \frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{1}{t}\right)}{dx/dt} \)
\( \displaystyle =\frac{-\frac{1}{t^2}}{2at} \)
\( \displaystyle =-\frac{1}{2at^3} \)
At t = 3:
\( \displaystyle -\frac{1}{54a}+\frac{1}{3}=0 \)
\( \displaystyle \frac{1}{54a}=\frac{1}{3} \)
\( \displaystyle a=\frac{3}{54}=\frac{1}{18} \)
Tip: The second derivative in parametric form is not \( \frac{d^2y/dt^2}{d^2x/dt^2} \); always divide \( \frac{d}{dt}\left(\frac{dy}{dx}\right) \) by \( \frac{dx}{dt} \).